Analytic Arithmetic in Algebraic Number Fields by Baruch Z. Moroz

By Baruch Z. Moroz

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Example text

195]. Re u I _< ~. and (12) 2q 6 ZZ, while in (3) and q • {O,1}. I I a = - ~, b = ~ If ( + i t ) Therefore 4 that F(s+1) Let Re u whenever I F ( ' q + u ) - l F ' c {/ + l - 2u ) 1 2 In for hold: ,q+u. -1 1 )F( 2 J I < (~ll+ul) 2 u b-a case of T h e o r e m The f o l l o w i n g . ~ with then if(u) I < (AI~+ul~) Proof. t E]R whenever (12) and e s t i m a t e (10) with and the f u n c t i o n a l = sF(s), I q ~ ~ follow equation s • ~. We h a v e f(- +it) = F (+it)r(~-it), from the p r o p e r t i e s t • ~+.

Is-~ I = 9/4 (15) shows is contained in the strip (15) a ~ Re s ~ b, that 9~4 ~(u) u 0 du = O(log ~(T+9/4)). (16) On the other hand, (/~) log - -9 = 4/£ It follows from 9/4 94/ ~ (/5) d__uu < ~ u- (16) and V(/~) (17) f ~ (u) o u du. og (18) ~(T+3)). ition 2. Lemma Estimate 3. Let T > O, < Re s < I, O < Im s < T, L(s,x) X 6 gr(k). = N(X,T) from Then = O}. Then + O(log(a(x)b(x)(3+T)n)). (19) follows X 6 gr(k). + ~(/~), (18). for card{sIO Let N(x,T+I) Proof. 30) in view of lemma 2. 56 --(s,x) = ~ (s-P) -1 I - I g(x)(s+s--Z~ -) (20) + O(A(X,t)), It-y1<1 I - ~ < Re where (3+Itl) n) Res< s < and p 2, t:= ranges Im s, over y:= the Im p, A ( x , t ) : = zeros of L(s,x) log(a(x)b(x)" in the s t r i p 0 < I.

S > ~ , it [78], 401, p. 2) )(log the Hadamard's that r3 -I ~11 ) (5) (3), M I = O(i (6) log(2+Oo(t))). Since O < a -<- 2(I-o) it follows from (5) and + (6) 2~(t) + 2(4~o(t)-I)-I g that -I 2 (I-~)+C2~o(t) M 2 = O(£(log Let 1 R = ~ o (t)- - i--=(1+~(t)) - Oo(t))M3 and let ). 2 in [78, p. 383], g(s)) on the c i r c l e IS-Sol = R. we have (8) M 3 <_ g(s O) (1+qo(t) 2) + 4Mqo(t) Conditions (2) and (3) give: g(s O) = O(Z) and M = O(log[B(f) and t h e r e f o r e (4) follows Cgrollar~ In notations I ,.